# Thread: Trigonometric Substitution

1. ## Trigonometric Substitution

integrate:$\displaystyle \frac{x^2}{\sqrt{4-9x^2}}$

i worked it down to this, first off is this right so far. And how would i simplify this so that its integrable. also how do you do the integrate sign in latex? thx guys.

integrate:$\displaystyle \frac{x^2}{\sqrt{4-9x^2}}$ = integrate:$\displaystyle \frac{\frac{2}{3}(cost)^2}{2sint} * (-\frac{2}{3} sint) dt$

2. Originally Posted by x5pyd3rx
integrate:$\displaystyle \frac{x^2}{\sqrt{4-9x^2}}$

i worked it down to this, first off is this right so far. And how would i simplify this so that its integrable. also how do you do the integrate sign in latex? thx guys.

integrate:$\displaystyle \frac{x^2}{\sqrt{4-9x^2}}$ = integrate:$\displaystyle \frac{\frac{2}{3}(cost)^2}{2sint} * (-\frac{2}{3} sint) dt$
I'm guessing you set $\displaystyle x = \frac{2}{3}\cos(t)$ yes?

That should be fine (although personally I would have chosen sin since it's derivative is positive...

But anyway...

$\displaystyle x = \frac{2}{3}\cos(t)$

$\displaystyle dx = -\frac{2}{3}\sin(t)dt$

Integral becomes...

$\displaystyle \int -\frac{2}{3}\frac{\tfrac{4}{9}\cos^2(t)}{\sqrt{4-4\cos^2(t)}}\sin(t) dt$.

$\displaystyle = \int -\frac{8\cos^2(t)}{54\sqrt{1-\cos^2(t)}}\sin(t) dt$

$\displaystyle = \int -\frac{4\cos^2(t)}{27\sin(t)}\sin(t) dt$

$\displaystyle = -\frac{4}{27} \int \cos^2(t) dt$

Solve from here, sub back in what t equals after you've integrated.

3. Originally Posted by x5pyd3rx
integrate:$\displaystyle \frac{x^2}{\sqrt{4-9x^2}}$

i worked it down to this, first off is this right so far. And how would i simplify this so that its integrable. also how do you do the integrate sign in latex? thx guys.

integrate:$\displaystyle \frac{x^2}{\sqrt{4-9x^2}}$ = integrate:$\displaystyle \frac{\frac{2}{3}(cost)^2}{2sint} * (-\frac{2}{3} sint) dt$

$\displaystyle \frac{x^2}{\sqrt{4-9x^2}}=\frac{x^2}{2\sqrt{1-\left(\frac{3x}{2}\right)^2}}$ , so now substittute $\displaystyle \frac{3}{2}\,x=\sin u\Longrightarrow \frac{3}{2}\,dx=\cos u\,du$ , and:

$\displaystyle \int \frac{x^2}{\sqrt{4-9x^2}}\,dx=\int\frac{x^2}{2\sqrt{1-\left(\frac{3x}{2}\right)^2}}\,dx$ $\displaystyle =\frac{1}{2}\int\frac{4/9\,\,\sin^2u\,\,(2/3)\,\cos udu}{\cos u}=\frac{4}{27}\int \sin^2u\,du$ $\displaystyle =\frac{2}{27}\,(u-\sin u\cos u)+C$

If you want now put back everything in terms of u instead of x.

Tonio

4. Thx so much guys you both where a bunch of help